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Thread: slope of tangent to curve

  1. #1
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    slope of tangent to curve

    find the slope of tangent to the curve x^2 +y^3=x-2 at the point (3,2)

    so i want to get it in the form of f(x)=y

    so y=(-x^2 +x-2)^1/3

    so i then differentiated this using the chain rule and then i subbed in 3 for x and i thought i would get the right answer but no
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  2. #2
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    Re: slope of tangent to curve

    $y' = \dfrac{1}{3}\left(-x^2+x-2\right)^{-2/3}(-2x+1)$
    $y'(3) = \dfrac{-5}{3(-8)^{2/3}} = \dfrac{-5}{12}$

    Let's try by implicit differentiation:

    $2x+3y^2\dfrac{dy}{dx} = 1$

    $\dfrac{dy}{dx} = \dfrac{1-2x}{3y^2}$

    Evaluate at $(3,2)$:

    $\dfrac{dy}{dx} = \dfrac{1-2(3)}{3(2)^2} = \dfrac{-5}{12}$

    I get the same answer from normal differentiation and implicit differentiation. What did you get?
    Thanks from markosheehan
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  3. #3
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    Re: slope of tangent to curve

    thanks. i was doing it wrong
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